One issue regarding the impact of voltage deviation on equipment
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Dear marine engineers, on page 9 of Liu Jiecai’s book \"Factory Power Supply\", it is stated as follows: \"The impact of voltage deviation on induction motors: When the terminal voltage of an induction motor is 10% lower than its rated voltage, since torque is proportional to the square of the terminal voltage, the actual torque will be only 81% of the rated torque. Meanwhile, the load current will increase by more than 5% to 10%, the temperature rise will increase by more than 10% to 15%, and the rate of insulation aging will double compared to the specified levels, thereby shortening the motor’s lifespan.\" When its terminal voltage is high, the load current and temperature rise will also increase, thereby damaging the insulation; this is detrimental to the motor as well and shortens its lifespan. ” I checked the mechanical characteristics of the motor; among them is the formula P=mUIcosa (where cosα is a factor related to the angle), and based on this formula, it makes sense that a low terminal voltage occurs ； But when the terminal voltage is high, the load current and temperature rise also increase, which is a bit hard to understand. Please ask an expert to clarify.
This post was last edited by jjli618 on 2010-7-1 21:52. Capacitive reactance is denoted by XC, capacitance by C (in F), and frequency by f (in Hz); therefore, Xc = 1/2πfc. The unit of capacitive reactance is ohms. Knowing the frequency f of the alternating current and the capacitance C, the capacitive reactance can be calculated using the above formula. Inductive reactance is denoted by XL, inductance by L (H), and frequency by f (Hz); therefore, XL = 2πfL. The unit of inductive reactance is ohms. Knowing the frequency f of the alternating current and the inductance L of the coil, the inductive reactance can be calculated using the above formula. Inductance: L = N × N × μ × A/l, where N is the number of turns, μ is the magnetic permeability of the core, A is the cross-sectional area, and l is the length of the magnetic path. The distributed capacitance of a circular coil is given by C = K * dielectric constant * 10^-6. It can be seen that impedance is related to factors such as frequency and material, and has no direct relation to voltage; once the motor is manufactured, its impedance remains constant. If the load impedance remains constant and the voltage exceeds the rated value, the current will increase. An increase in current leads to a nonlinear increase in flux, causing the stator core of the motor to heat up. As a result of these interactions, the motor current rises under certain conditions, and the temperature increases. An increase in temperature is a test for insulation; when the temperature rise exceeds the rated level, it damages the insulating material and reduces its insulation capacity. Reduces the service life of the motor. Of the two situations you mentioned, in the first case the power supply voltage is low; to overcome this, more effort is exerted to generate current, thereby increasing torque. In the latter case, the supply voltage increases; with a constant impedance, this leads to an increase in current, and it is then that the temperature rises.
Reply to 3# jjli618: The answer is indeed very clear, thank you!
The explanation on the third floor is quite comprehensive: when the voltage increases, the increase in eddy currents in the stator and rotor cores leads to an increase in the temperature of those cores.